# Polar Curve and maximum width

Find the maximum width of the petal of the four-leaved rose $r = \cos2\theta$, which lies along the x-axis

Here is the solution

Can someone tell me how on earth did the solution come up with the first step?

The statement "The maximum width of the petal of the rose which lies on along the x-axis is twice the largest y-value of the curve on the interval..."

Maybe the curve I sketched wasn't great, but I did it again on Mathematica and I still couldn't see how they notice this "ingenuous" subtle observation.

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$r = \cos\theta$ is a circle; did you forget to include part of the equation for the four-leaved rose? –  amWhy Dec 27 '12 at 21:03
Woops, I deleted a $2@ – Hawk Dec 27 '12 at 21:04 Now, that makes more sense! ;-) – amWhy Dec 27 '12 at 21:07 ## 3 Answers The petal lies along the positive$x$-axis. To say it differently, the positive$x$-axis is the axis of symmetry of the the petal. The length of the petal is therefore measured horizontally, from the origin to the tip of the petal. The width at any point is the size of the petal measured perpendicularly to the axis, i.e., at right angles to the$x$-axis. Thus, the width at any point is measured parallel to the$y$-axis and is therefore the distance between the top and bottom of the petal at that value of$x$. Since the petal is symmetric about the$x$-axis, that width is twice the$y$-value of the top edge of the petal. This reaches its maximum when the$y$-coordinate of the top edge of the petal is as large as possible. - Oh I see what you mean. Thanks – Hawk Dec 27 '12 at 21:17 @sizz: You’re welcome. – Brian M. Scott Dec 27 '12 at 21:20 Perhaps this image will help: it's the graph of the petal which lies long the positive x-axis, on the interval$0 \le \theta \le \dfrac{\pi}{4}$: $$\quad$$ $$\quad$$ So, "The maximum width of the petal of the rose which lies on along the x-axis is twice the largest y-value of the curve on the interval..." Note: one could have used the same procedure outlined above by considering the maximum y-value for$\dfrac{3\pi}{4} \le \theta \le \pi$, and get the same end-result. $$\quad$$ So there's nothing particularly special about the interval$0 \le \theta \le \pi/4$: the same maximum width or each petal can be determined using any one of eight possible intervals. It just happens that choosing the interval$\theta \in (0, \pi/4)\$ simplifies things a tad.

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Not an answer, but a picture helps...

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